How Do Subfields of Finite Fields Relate to Their Elements?

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A finite field with p^n elements contains exactly one subfield with p^m elements for each divisor m of n, as the degree of the subfield must divide the degree of the larger field. The polynomial Φ_p(x) = (x^p - 1)/(x - 1) is irreducible over the rationals for any prime p, and its roots can be shown to be distinct by considering the powers of a root α. Specifically, if α is a root, the distinct elements {α, α^2, ..., α^(p-1)} can be derived from the properties of roots of unity, where α^p = 1. The discussion emphasizes the relationship between field extensions and the uniqueness of subfields based on their sizes.
Mathmajor2010
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Homework Statement


Show that a finite field of p^n elements has exactly one subfield of p^m elements for each m that divides n.

Homework Equations


If F \subset E \subset K are field extensions of F, then [K:F] = [E:F][K:F] . Also, a field extension over a finite field of p elements has p^n elements, where n is the number of basis vectors in the extension field.

The Attempt at a Solution


We have a field K of p^n elements, which has degree n because it has n basis vectors. Then, if we have a field extension E of F s.t. F \subset E \subset K, then the degree of E has to divide K. That is, the degree of E, call it m, has to divide n. Therefore, E has p^m elements, where m divides it. Does this sound right?Another question I'm stuck on is the following:

Homework Statement


Let \Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1. This polynomial is irreducible over the rationals for every prime p. Let \alpha be a zero of \Phi_p . Show that the set \{\alpha, \alpha^2, ... \alpha^{p-1} are distinct roots of \Phi_p .

The Attempt at a Solution


I'm not really sure how to approach this. It seems to me that since \alpha is a root, we can find an extension field. So, Q(\alpha) is spanned by 1, \alpha, \alpha^2, ... \alpha^p-2. But, I don't see how to show that each of these is a root. Any suggestions of hints for this one? Thanks!
 
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Mathmajor2010 said:

The Attempt at a Solution


We have a field K of p^n elements, which has degree n because it has n basis vectors. Then, if we have a field extension E of F s.t. F \subset E \subset K, then the degree of E has to divide K. That is, the degree of E, call it m, has to divide n. Therefore, E has p^m elements, where m divides it. Does this sound right?

You have shown neither that E exists nor that it is unique. All that you've said is that subfields have size a power of p.

Another question I'm stuck on is the following:

Homework Statement


Let \Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1. This polynomial is irreducible over the rationals for every prime p. Let \alpha be a zero of \Phi_p . Show that the set \{\alpha, \alpha^2, ... \alpha^{p-1} are distinct roots of \Phi_p .


Have you tried just plugging in \alpha^2 and see what you get? You'll get a sum of a bunch of different values for \alpha. Keep in mind that \alpha is a root of unity, so you know for example that \alpha^p=1. See if you can use this to rewrite those powers of \alpha
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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